As shown in the figure, in △ ABC, ab = AC, D, e are the points on BC and AC respectively, and de ‖ AB, EA = ed. please explain that ad bisects BC vertically

As shown in the figure, in △ ABC, ab = AC, D, e are the points on BC and AC respectively, and de ‖ AB, EA = ed. please explain that ad bisects BC vertically

It is proved that: as shown in the figure, ∵ EA = ed, ∵ de ∥ AB, ∵ de ∥ AB, ∵ 1 = ∵ 3, ∵ 1 = ∵ 2. That is, ad bisects ∵ BAC. And ∵ AB = AC, ∵ ad is the middle vertical line of side BC, that is, ad bisects BC vertically

Given the function f (x) = e to the power of x-kx, X belongs to R
If k = e, try to determine the monotone interval of function f (x)
2 if k > 0 and f (absolute value x) > 0 is constant for any x belonging to R, try to determine the value range of real number K
Let f (x) = f (x) + F (- x), prove: F (1) f (2) Half n of F (n) > (n + 1 power of E + 2) (n belongs to n *)

So when x ＞ 1, it increases monotonically and when x ＜ 1, it decreases monotonically
2. It can be seen from the question that the function is symmetric about the y-axis, so the first quadrant analysis has e ~ x > KX, and the derivation has e ~ x > K. because x is in the first quadrant, the value of E ~ x is greater than or equal to 1, so the value range of K is 0 < K < 1
3. There's nothing I can do. I can't understand the title

If the two right sides of a right triangle are 5 and 12, what is the radius of its circumcircle

According to Pythagorean theorem
The hypotenuse of a right triangle is the root (5 * 5) + (12 * 12) = 13
The hypotenuse of a right triangle, even the diameter of its circumscribed circle (this is the law)
So his radius is 13 / 2 = 6.5

Given the complete set u = R, set a = {x | x ＜ a}, B = {x | 1 ＜ x ＜ 2}, and a ∪ (cub) = R, find the value range of real number a

∪ cub = {x | x ≥ 2 or X ≤ 1}, and a ∪ (cub) = R
∴﹛x|1﹤x﹤2﹜⊆A
∴2≤a
That is, a ≥ 2

In △ ABC, ∠ C = 90 °, AC = BC, D is the point on AC, AE ⊥ BC intersects the extension line of BD at e, and AE = half BD, DF is perpendicular to ab at F

When the extension line of AE intersects with the extension line of BC at the point G, it is known that: AC = BC. ∠ ACG = ∠ BCD = 90 °≌ △ ACG ≌ △ BCD, so BD = AG. It is also known that: BD = 2ae; then Ag = 2ae, so AE = eg

F (x) = x & # 178; + 2aX + A & # 178; + B, if x ∈ R, f (x) ≥ 0, then the value range of B
② If f (x) is an even function, then a =?

① It requires that Δ = (- 2A) &# 178; - 4 (a) &# 178; + b) = - 4B ≤ 0, so B ≥ 0
②f(-x)=(-x)²+2a*(-x)+a²+b=x²-2ax+a²+b
Even function, then f (- x) = f (x), so - 2aX = 2aX, so a = 0

As shown in the figure, in △ ABC, make the bisector ad of ∠ BAC through C, the perpendicular foot is D, ad intersects BC at g, de ‖ AB intersects AC at E. (1) verification: AE = CE; (2) make the bisector CF of ∠ BCA intersects ad at P, intersects AB at F, verification: PCD = 12 ∠ B; (3) under the condition of (2), if ∠ B = 60 °, verification: AF + GC = AC

It is proved that: (1) extending CD, the extension line of intersection AB is at h, ∵ ad ⊥ ch, i.e., ∵ ADC = ∵ ADH = 90 °, had = ∵ CAD, ∵ H = ∵ ACh, ∵ ah = AC, i.e., ∵ ACh is isosceles triangle,